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Question:

If $\lambda_{1}$ and $\lambda_{2}$ are the wavelengths of the third member of Lyman and first

member of the Paschen series respectively, then the value of $\lambda_{1}: \lambda_{2}$ is :

1. (1) $1: 3$

2. (2) $1: 9$

3. (3) $7: 135$

4. (4) 7:108

Correct Option: , 3

Solution:

(3)

For Lyman series

$n_{1}=1, \quad n_{2}=4$

$\frac{1}{\lambda_{1}}=R z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$

$\frac{1}{\lambda_{1}}=\mathrm{Rz}^{2}\left(\frac{1}{1_{1}^{2}}-\frac{1}{4^{2}}\right)$

$\frac{1}{\lambda_{1}}=\frac{15 \mathrm{Rz}^{2}}{16}$

$\lambda_{1}=\frac{16}{15 \mathrm{Rz}^{2}}$

For paschen series

$n_{1}=3, \quad n_{2}=4$

$\frac{1}{\lambda_{2}}=\mathrm{Rz}^{2}\left(\frac{1}{3^{2}}-\frac{1}{4^{2}}\right)$

$\frac{1}{\lambda_{2}}=R z^{2}\left(\frac{16-9}{9 \times 16}\right)$

$\frac{1}{\lambda_{2}}=\mathrm{Rz}^{2}\left(\frac{7}{9 \times 16}\right)$

$\lambda_{2}=\frac{9 \times 16}{7 \mathrm{Rz}^{2}}$

So,

$\frac{\lambda_{1}}{\lambda_{2}}=\frac{\frac{16}{15 \mathrm{Rx}^{2}}}{\frac{9 \times 16}{7 \mathrm{Rx}^{2}}}$

$=\frac{16 \times 7}{15 \times 9 \times 16}$

$=\frac{7}{135}$